Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = -\dfrac{2}{3}x - 1}\enspace$ and passes through the point ${(-5, -6)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${-\dfrac{2}{3}}$ , and its negative reciprocal is ${\dfrac{3}{2}}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = \dfrac{3}{2}x + b}\enspace$ We can plug our point, $(-5, -6)$ , into this equation to solve for ${b}$ , the y-intercept. $-6 = {\dfrac{3}{2}}(-5) + {b}$ $-6 = -\dfrac{15}{2} + {b}$ $-6 + \dfrac{15}{2} = {b} = \dfrac{3}{2}$ The equation of the perpendicular line is $\enspace {y = \dfrac{3}{2}x + \dfrac{3}{2}}\enspace$. ${m = \dfrac{3}{2}, \enspace b = \dfrac{3}{2}}$